The correct answer is $\frac{{{\text{W}}{{\text{L}}^3}}}{{8{\text{E}}I}}$.
The maximum deflection of a simply supported beam of span $L$, carrying an isolated load at the center of the span, can be calculated using the following formula:
$$\delta = \frac{{{\text{W}}{{\text{L}}^3}}}{{8{\text{E}}I}}$$
where:
- $\delta$ is the maximum deflection,
- $W$ is the load,
- $L$ is the span,
- $E$ is the Young’s modulus, and
- $I$ is the moment of inertia.
The formula can be derived by considering the bending moment diagram for a simply supported beam with an isolated load at the center of the span. The bending moment diagram is a plot of the bending moment along the length of the beam. The maximum bending moment occurs at the center of the span, and is equal to $WL/2$.
The deflection of a beam can be calculated using the following formula:
$$\delta = \int_{0}^{L} \frac{M(x)}{EI} dx$$
where:
- $\delta$ is the deflection,
- $M(x)$ is the bending moment at a distance $x$ from the left end of the beam,
- $E$ is the Young’s modulus, and
- $I$ is the moment of inertia.
The bending moment at a distance $x$ from the left end of the beam is equal to $WL/2$ for $0 < x < L/2$, and is equal to 0 for $L/2 < x < L$. Substituting these values into the formula for the deflection gives:
$$\delta = \int_{0}^{L/2} \frac{WL}{2EI} dx + \int_{L/2}^{L} \frac{0}{EI} dx = \frac{WL^3}{16EI}$$
The maximum deflection occurs at the center of the span, where $x=L/2$. Substituting this value into the formula for the deflection gives:
$$\delta = \frac{WL^3}{16EI} = \frac{{{\text{W}}{{\text{L}}^3}}}{{8{\text{E}}I}}$$
Therefore, the maximum deflection of a simply supported beam of span $L$, carrying an isolated load at the center of the span, is $\frac{{{\text{W}}{{\text{L}}^3}}}{{8{\text{E}}I}}$.